How to calculate the standard deviation and mean of a group of individual statistics

Is there a formula to calculate the standard deviation of a group or individual statistics

Example in a sporting fantasy league I can calculate the summary statistics of any one player.

Example

one player has a Mean 75, Standard deviation 26.

But what if I selected a group of 18 players with the same summary stats, can I estimate the summary statistics of this group?

The Mean I believe can just be multiplied 1350 (75*18),

but the standard deviation isn't as simple as using a multiplication?

To add to this- if the group of 18 players each had unique summary stats, can the groups summary stats also be explained/calculated? example

Player 1 - mean 75 sd 26 Player 2 - mean 100 sd 28 Player 3 - mean 105 sd 25 . . Player 18 - mean 72 sd 23

Any help would be great.

Thanks heaps

• Means add up, so if player $$A$$ has mean $$\mu_a$$ and player $$B$$ has mean $$\mu_b$$, the mean of their sum is $$\mu_a+\mu_b$$. You can therefore multiply your common mean by 18 (because that's just adding up the 18 means)
• Standard deviations don't add up, but their squares (variances) do! So, if standard deviation for player A is $$\sigma_a$$ and standard deviation for player B is $$\sigma_b$$, the standard deviation for their sum is $$\sqrt{\sigma_a^2 + \sigma_b^2}$$. In your case, square the standard deviations to get variances, add them up (multiply by 18) and then take the square root to have a standard deviation rather than a variance again
• Very good question! Yes, you can! If $X$ follows a $N(\mu_x, \sigma_x^2)$, Y follows $N(\mu_y, \sigma_y^2)$ and $X$ is independent from $Y$, then $X+Y$ follows a $N(\mu_x + \mu_y, \sigma_x^2 + \sigma_y^2)$ (see en.wikipedia.org/wiki/…) so these properties still apply even if means are different – David Jun 17 at 11:39